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00:00 - 19:0019:00 - 00:00

00:03
@DonLarynx Yes, but I don't see what you want to do.
I want to show that $A | B$ can't be a separation of $K$
by showing it is connected
I know it is compact
because closed intersection is closed and a closed subset of a compact set is compact.
@DonLarynx Showing what is connected?
$K$. I assume there exists a separation $A | B$ of $K$ and show it's a contradiction
@DonLarynx Yes, that is the point of the proof. Did you read the proof in that question you link to?
Look, I know $K$ is compact. But I want to show there exists finite subcovers of $A$ and $B$. I know that $A \cup B$ has a finite subcover.
I've spent hours on this @Pedro, don't insult me
That's assuming it's Hausdorff.
00:09
@DonLarynx I am not insulting you, I am just wondering if you've read that proof. If there is something you don't understand, I can explain.
I know $K$ is compact. But I want to show there exists finite subcovers of $A$ and $B$. I know that $A \cup B$ has a finite subcover.
@Pedro okay, sorry.
I can't use that proof
because the space may not be Hausdorff
@DonLarynx I am not sure the claim is true in general.
You can ask that on main, or maybe @DanielFischer knows.
It is true in general because my sadistic teacher assigned it
Please help me
lol
What is the claim?
4
Q: Connectedness and intersection problem

phoenixLet $X$ be a compact metric space, and let $K_1,K_2,\dots$ be a sequence of subsets of $X$ which are nonempty, closed, and connected, and which satisfy $K_{n+1}\subset K_n$ for every positive integer $n$. Prove that intersection of $K_n$ for $n$ from $1$ to infinity is connected. So, this was c...

@Daniel
00:13
@DanielFischer Sorry. =)
NO!
connected
I already know it is compact
@DonLarynx You only need compactness, you don't need a separation axiom like the Hausdorff axiom.
@DonLarynx Can you copy the assignment verbatim?
@Pedro the problem is exactly the same, but if you insist.
@DonLarynx Then, why doesn't the answer apply?
00:16
@DonLarynx But $X$ is a compact topological space?
Once again
it is straightforward it is compact
@DonLarynx I stick to my remark.
The answer given in that question works here since the conditions are exactly the same.
@DonLarynx What does $N_{1/n}(A)$ denote?
(Nota that $X$ being compact is irrelephant.)
The professor did not define it and we are all confused @pedro.
me and my colleagues
@DonLarynx I am pretty sure the prof. has defined it. Look in your book or notes.
00:18
@DonLarynx Ten internet points that $N_{1/n}(A)$ is the $1/n$-neighbourhood of $A$.
My colleague thinks $N_{\frac{1}{n}}(A) = \bigcup\{N_{\frac{1}{n}}(a) : a \in A\}$
@DanielFischer A net?
That is, $N_{r}(A) = \{ x : \operatorname{dist}(x,A) < r\}$.
Which is what Don's colleague thinks.
@Pedro: I am surrounded by 4 colleagues each with $A$'s in the class. it is not the case he defined it.
@DonLarynx Having an $A$ is not a free ticket to anything, you know. =)
Anyways, 5 minds should be able to write something up!
Right?
You have a long hint.
Have a blast.
It will be better for you if you work it by yourselves.
And more fun that talking with random strangers on the internet.
00:21
ummm we've been working on it since 3 Pm + friday + saturday....
sorry bro
Really?
Have you followed the hints?
Can you show the first thing they ask you to show?
That for each $n$ we have $N_{1/n}(A)\cap N_{1/n}(B)\cap C_n\neq\varnothing$?
Here is what we have so far.
following from the picture.
I'll write "done" when I am finished
$C \subset U \cup V$ and $U \cap V = \emptyset$
@DonLarynx Words, words.
$F_n := C_n \ \{U \cup V\}$
I like words.
00:26
C is the subset of two open sets $U$ and $V$
and these two open sets are disjoint
now take a sequence of compact sets $\{F_n\}_{n = 1}^{\infty}$
@DonLarynx You were going to prove $N_{1/n}(A)\cap N_{1/n}(B)\cap C_n\neq\varnothing$.
and set it equal to $C_n \setminus \{U \cup V\}$
This intersection empty implies that one of the finite intersections are empty.
> The current answers do not contain enough detail.
Nice bounty reason, the question has zero answers to date ;)
lol I should try that with some other vacuous truth.
@DanielFischer I wonder. This is probably easy, I just haven't thought about it. One can define a polynomial in $R[x_1,\ldots,x_r]$ to be homogeneous if all its constituent monomials have the same total degree. But we can also define them as those for which $P(t{\bf x})=t^p P({\bf x})$ for some constant $p$ (the total degree of the monomials). The former implies the latter easily, can we show the latter implies the former without much problem?
00:38
@PedroTamaroff In $\mathbb{R}$ or $\mathbb{C}$, without much problem. Not sure if there may be a problem in some weird rings.
@DanielFischer Ah, I see.
@PedroTamaroff This works for any integral domain, I am sure.
I am reading the proof that any polynomial can be written as a sum of symmetric polynomials.
And it has some nifty moves. Or not.
Maybe I should let @anon judge? =)
@KarlKronenfeld Fire away.
@PedroTamaroff It is impossible to define degree when $R$ is not an integral domain.
letting $t$ be a formal variable it is of course true
00:39
So this is the best you can do.
@anon Aha.
but the equality as polynomial functions probably need not hold over finite rings
@PedroTamaroff Hm, monomial degree is still well-defined. My bad.
some sources use lexicographic reasoning to prove fund thm sym polys, but I prefer this approach
@PedroTamaroff There are two kinds of "false-positives" that can occur using the equation $P(t\mathbf x)=t^pP(\mathbf x)$. One is based on the fact that there could conceivably exist a number $n$ such that $a^n=a$ for all $a\in R$. The other is that you can lose information going from a formal polynomial in $R[\mathbf x]$ to a polynomial function. In particular my guess that it would work for all integral domains $R$ is wrong.
leo
leo
01:00
@KarlKronenfeld but the equality proposed as homogeneity must holds for all $t$, not just particular ones
If I understand correctly
@anon why not?
oops, I mean the equality as functions in each variable probably need not imply P() is homog
@leo Yes, that is correct.
leo
leo
oh I see. Then, probably
I think any finite field fails on both counts and an arbitrary finite ring fails on at least the second point I mentioned.
leo
leo
$xy - y$ over $\Bbb F_2$
01:09
Yeah, any polynomial over $\mathbb F_2$ will pass that test.
leo
leo
and the monomials have not the same total degree
x+x^3 over F_3
leo
leo
yep
I learned the hard way that dishwasher soap is not edible.
leo
leo
ha ha ha
at least taste bad
01:15
I had a horrible allergic reaction for approx. 45 min.
I poured some water into a drinking glass, and I noticed suds. Since I am an idiot, I drank the water anyway.
leo
leo
:(
01:35
Problem: Evaluate $\zeta(4)$ using the Poisson summation formula. I'm thinking of finding the Fourier transform of $\frac{1}{(x^2 + a^2)^2}$ by convolving the transforms of $\frac{1}{x^2 + a^2}$ and then taking the limit as $a \to 0$. The sum over the convolved Fourier transforms should be some a sum over some kind of exponential which I'm hoping can be massaged into something trigonometric
Here's how it's done for $\zeta(2)$ (aside from the inexplicable error at the very end):richardtroll.files.wordpress.com/2010/11/zeta-21.pdf
02:02
Is there a way to sort starred chat posts by number of stars?
data.stackexchange.com doesn't access chat posts, so I can't find a workaround that way, either.
@MarkS. What's funny is that the list of messages starred by me is sorted in descending order of number of stars.
@KarlKronenfeld That's...something.
Does this request go on meta.stackoverflow?
02:34
meta.stackoverflow appears to have gone down as I was searching to see if there was already a request there
@usukidoll what're you working on?
nothing
More Laplace transforms?
$\varnothing$ ?!?!
Why does $\varnothing$ look so much nicer than $\emptyset$? shouldn't "emptyset" be the nicest symbol for that?
02:40
$\varnothing$_$\varnothing$
done with my homework
just chilling
after meta.stackoverflow started throwing errors, now math.SE says "This site is currently in read-only mode; we’ll return with full functionality soon." Looks like this is the end.
@usukidoll do you know what a ring is?
no
D:
I'm a sad math major I know D:
It's a set with similar algebraic properties to the integers, except it doesn't need to be commutative under mult.
do you get it now?
So in a ring a(b + c) = ab + ac, just like in $\Bbb{Z}$.
$\Bbb{Z}$ is an example of a ring
also said to form a ring
02:44
Anyone else having problems with the site?
@EnjoysMath Just out of curiosity, are rings related to something you and usukidoll were discussing before, or did you just think "hey, I'm going to tell someone what a ring is now"? I mean this in the best possible way.
the latter, why?
"just out of curiosity"
I was going to move onto modules but looks like I lost them
at rings
"just chilling" made me think of chilling in CGT, but that's a little harder to jump into an explanation about than rings.
02:49
@MarkS. Did you know that $\Bbb{Z}/p^k \Bbb{Z} \otimes_{\Bbb{Z}} A \approx A(p)$, where $A$ is a finite abelian group and $A(p)$ is it's Sylow $p$-subgroup?
@MarkS. I learned that today \o/
for big enough k
I don't think I ever learned that, but I bet it doesn't work for all $k$. Beaten.
ah, you're right
*don't think I ever learned that explicitly
It's isomorphic to $Z/p^k Z \otimes A(p)$ where $A(p)$ is the subgroup of all $p$-power order elements.
since $A = \bigoplus_{p} A(p)$
and tensor product distributes over dir sum
and you can show that $Z/p^k Z \otimes A(q) = 0$ for $\gcd(p,q) = 1$.
02:54
That sounds about right, yes. It's been a while for me.
If $S_n=\left[-\frac{n}{n+1},\frac{n}{n+1}\right]$, then why union of $S_n$ is (-1,1) and not [-1,1] ?
Is $1$ in any of those intervals? If so, for which $n$?
@MarkS. Thank you for so quick reply.
@Sush Sure thing. If that didn't make it clear, let us know.
@MarkS. Can we use this same argument as proof?
03:03
Proof that 1 is not in the union? Basically yes. But it's not proof that the union contains all of (-1,1).
@Sush Some advice for things like math.stackexchange.com/questions/578990 . If you're not satisfied with an answer, you can upvote it but then not accept it, so that other people know the question hasn't been answered to your satisfaction.
(or not upvote it if you don't think it's worth it)
@Sush However, you should also try to follow the guidelines found here: meta.math.stackexchange.com/a/1804/26369 if this is a homework question/textbook question. Explain what your thoughts have been in detail, etc.
@MarkS. Sorry. Now I have made it unaccepted. I will surely keep your advice in my mind for later questions.(I am not native English speaker, so please pardon me if I am wrong in communication skill. ) This was not any homework, but I was self reading Mathmatics for Economists by Simon and Blume.
Don't worry about your English. What you've said has been clear enough. If you edit your question to say "I was self reading Mathmatics for Economists by Simon and Blume." it will help people write answers close to your level/style.
@MarkS. Thanks a lot,Sir. Just going and editing.
03:57
That's what I get for misreading a question
what is that
@Ethan Mean answer: Figure it out. Nice answer: $(t-2) \sin \left(\frac{2 \pi t}{t-2}\right)$
oh, ok
What about $\zeta$?
I should probably have written $(2-t)\sin\left(\frac{2\pi t}{2-t}\right)$ since $t\in [0,2]$
@MarkS. That looks like the solution to the differential equation $y'' + e^{x}y = 0$
04:12
@bitrex But those bessel functions have a curved envelope, whereas this just has boring straight lines.
ah you're right
It looked kinda curved to me but maybe I'm seeing things
I just put a straightedge to the screen and yes I am seeing things :)
Well, there are some artifacts of quick computer graphing.
That's to scale
04:47
So, I'm trying to plan my lecture for determinants for a college algebra class, and it occurs to me that I still don't really understand what a determinant tells us, or some interesting applications of it. I could go with finding area of parallelograms/pipeds, but is there anything else that would possibly interest a college algebra class?
05:30
create a matrix to transform "amount of alcohol" into "number of down bitches"
haha nice one
@anon what is basis for $Z[x_1][x_2]$ as a module over $Z[x_1]$?
anyone?
I think it's $\{1, x_1, x_1^2, \dots \}$.
@anon
How can I be certain?
05:51
yes
{1,x,x^2,...} is an R-basis for R[x]
polynomials are by definition R-linear combinations of powers of x which are equal iff all coefficients are equal
well, actually you have the subscripts wrong
the powers of x2 are the basis for Z[x1][x2]
\o/ I'm working on your problem
...my problem?
06:14
@anon
0
A: Basis for $\Bbb Z[x_1,\cdots,x_n]$ over $\Bbb Z[e_1,\cdots,e_n]$

Enjoys MathProve the statement for the case $n = 1$. This will be the base case for an induction. Let $R = \Bbb{Z}[e_1, \dots, e_{n-1}], \ S = R[e_n]$. When refering to $R$ we're refering to the isomorphic copy of $R$ in $S$. So we have $R \lt S$ as rings. Let $M = \Bbb{Z}[x_1, \dots, x_{n-1}]$, and $N...

Please check. Kthx
And get back to me, I really want to know if that trick worked
feeding cats. afk
Morning!
Could you please give some upvote to the following?http://math.stackexchange.com/questions/556977/gaussian-integrals-over-a-half-space
Just to give it some attention! Thanks!
06:31
@anon yo problem, man
07:01
Hi. If anyone is good with regular language, please tell me for each one if those languages are regular or not (I think they're all irregular, and proved with pumping lemma):
$\{ a^ky \mid y \in \Sigma^* , |y|_a \leq k \}$
$\{ a^ky \mid y \in \Sigma^* , |y|_a \geq k \}$
$\{ 0^n1^k2^m \mid m=max(n,k) \}$
 
2 hours later…
09:18
is there a way to determine the highest frequency component in an arbitrary algebraic F(x)
09:51
I mean, I'm nearly sure there is by intuition, but I don't know who's theory to google for
10:25
@DanielFischer, any help please? You had a clever idea, but you left me in the dark! :) Could you give some more info about? Thanks a lot, really!
 
3 hours later…
13:26
Hello! Help me please, what is a definition of an OUTWARD pointing normal vector to a boundary point of a RIemann surface?
13:36
@Nimza, Does it help?
@nullgeppetto no, with surfaces it is clear what does it mean; but with Riemann surfaces it's not so obvious
@Nimza, :/ sorry..
ah, thanks anyway)
@Nimza, :)
14:23
@chris hi, Chris! :) whenever you can, I would like to discuss one or two exercises with you :D little geometry thing.
 
1 hour later…
15:48
@AlexanderGruber Hey.
What's the news?
16:00
I have lost focus.
Greetings
I was doing a totally different thing, and now I am all into the Bernoulli numbers.
@Chris'ssis Hi
@MatsGranvik Hello :-)
@Chris'ssis I have something for you. Find the value of $$\int_0^{\pi/2}\frac{x dx}{\tan x}$$
@PedroTamaroff This stuff is too easy to me.
16:02
@PedroTamaroff I will be back later.
:12338462
Sorry I pinged the wrong person.
I have to go.
@Chris'ssis So you already know the value?
@PedroTamaroff Mentally did it (start integrarting by parts, and then you get a well-well-known integral)
@Chris'ssis Heh. OK.
@PedroTamaroff there you can let $x=2y$ and then make use of the symmetry of the interval.
@Chris'ssis Can you solve it without integrating by parts?
16:05
@PedroTamaroff Yes.
@Chris'ssis OK?
@PedroTamaroff $$I(\alpha)= \int_0^{\pi/2}\frac{\arctan(\alpha \tan(x)) dx}{\tan x}$$
@Chris'ssis Right. =)
Too well known of an integral.
Did you see the other one?
@PedroTamaroff Which one?
$$\int_0^{\pi}\frac{xdt}{x^2+\sin^2t}$$
16:07
@PedroTamaroff No, I didn't see it.
@Chris'ssis Let me know what you get.
@PedroTamaroff OK
Hello every one I put the Question last night but no one answered it yet. Are there any way to help me? thank you
@Zeezee What is the question? (If you're going to link, use [text](link) )
this is link of my question
16:14
@Zeezee Sometimes it takes a while before someone who can answer your question sees it. If a few days go by and you still don't have an answer, you might think of adding a bounty.
@Zeezee You did not use [text](link).
first What is the bounty and link or text you said
@Zeezee I changed your comment to the proper format
I am totally new here, thank you
@PedroTamaroff Basically, the integral is elementary, but maybe you think of a fast way. You may firstly factorize $x^2$ in denominator. Then, by symmetry you only consider the integral over half of the interval.
16:16
@Chris'ssis That kind of integral is handled in celestial mechanics. Let me try.
@Chris'ssis The tan in the denominator may be a problem
@robjohn Now, I was talking about $$\int_0^{\pi}\frac{xdt}{x^2+\sin^2t}$$
@Chris'ssis Ah, that is different
@robjohn No, not at all.
@Chris'ssis What did you get?
@PedroTamaroff $$\frac{\pi}{\sqrt{x^2+1}}$$
16:26
@Chris'ssis Yes, but you're missing a factor of $1/2$.
@PedroTamaroff Do I?
@Chris'ssis Yep.
@Chris'ssis Anyways, how did you solve it?
Can anyone help me disprove that $L = \{ w \ mid |w|_a = 2^n + 273 , n \in \mathbb{N} \}$ over $\Sigma = \{a,b\}$ is non regular?
16:34
@PedroTamaroff By going the long way. Did you do it by some differentiation?
@Chris'ssis I haven't done it yet, but I was curious as to what way you chose. Can you explain?
@PedroTamaroff Well, there are more things you cna do here. I'm not sure yet what is the best one: One way is Weierstrass substitution
@Chris'ssis Ah, sure.
I have a far more challenging (I think?) integral, from Polya and Szego. Looks quite bananas.
@TheNotMe No idea.
@PedroTamaroff Another way is to multiply both numerator and denominator by $1/\sin(x)^2$. Then let $\cot(x)=t$.
@Chris'ssis Aha.
$$\int_0^{2\pi } {\frac{{{{\left( {1 + 2\cos t} \right)}^n}\cos nt}}{{1 - r - 2r\cos t}}dt} = \frac{{2\pi }}{{\sqrt {1 - 2r - 3{r^2}} }}{\left( {\frac{{1 - r - \sqrt {1 - 2r - 3{r^2}} }}{{2{r^2}}}} \right)^n}$$
$-1<r<1/3$
@robjohn Have fun $\uparrow$.
16:40
@PedroTamaroff Maybe it seems crazy, but I'd try this by induction is the right side is given.
@Chris'ssis Heh, not a bad move.
Wonder who was the person that obtained the general closed form.
@PedroTamaroff im not even able to use the pumping lemma to disprove
16:55
@Chris'ssis $u=\tan(x)$ makes it easy
@robjohn thank you for help me
@robjohn yeah
@Pedro thank you so much for your answer but I have problem yet
@PedroTamaroff hmmm, it may have some connections with Chebyshev polynomials
17:09
We miss a nice series near to this nice integral $$\sum_{n=1}^{\infty}\frac{\Gamma((2n+1)/n)}{n^2 \Gamma(n)}$$
:D
Suppose that $f(z)=z+a_2 z^2+a_3 z^3+ \dots$. Why then we can say that $$\frac{1}{f(z)}=\frac{1}{z}+c_0+c_1 z + c_2 z^2 + \dots$$ for some $c_i \in \mathbb{C}, i \in \mathbb{N}$ ?
Anybody? ;)
Maybe @DanielFischer :)
Hi.
$$\frac{1}{f(z)} = \frac{1}{z(1+a_2z + a_3z^2+\dotsb)} = \frac{1}{z}\left(1 - (a_2z+a_3z^2+\dotsb) + (a_2z+a_3z^2+\dotsb)^2 - (a_2z+a_3z^2+\dotsb)^3 +\dotsb\right)$$
@DanielFischer Oh, nice idea. Thank you :)
And I don't understand way nobody can help me with this my question: math.stackexchange.com/questions/578158/…. I was thinking that this is elementary, but.... I will give bounty if nobody answer.
17:52
Sorry, there is a typo above. It's $$\sum_{n=1}^{\infty}\frac{\Gamma((2n+1)/2)}{n^2 \Gamma(n)}$$
@Chris'ssis to which integral?
@robjohn the last one posted by @PedroTamaroff
@Chris'ssis Ah. Let me try that integral...
@Chris'ssis Hola, Chris
@Charlie Greetings! :D How are you doing?
18:05
@Chris'ssis I'm fine, I'm fine, and you?
@Charlie not too bad. Working on some proofs.
@Chris'ssis cool :)
@Charlie :D
@robjohn I wish you have luck!
@Charlie Não. Você não está fina, você está gôdinha.
:3
@GustavoBandeira -_- gôdinha é o ....
18:11
@Charlie Sêza educadinha. :3
@EnjoysMath You really have to let the arrows lead the way. There is not much you can do. Pick an element of $b'\in B'$. Where can you go? To $\phi'(b')=c\in C'$ by $\phi'$. But you know $\gamma$ is surjective, so you get back to $C$. And so on...
So it's follow the yellow brick road? Lemme try that
I mean, in each stage you don't have options, there is one thing you can do that will lead you somewhere: commutativity of the diagram or exactness.
(Well, and moving on through the arrows.)
18:19
@Charlie Cuidado, gôdinha. Eu faço bullying. :3
@GustavoBandeira só um segundo
@PedroTamaroff Can you look this. What's the meaning of show? Is the procedure indicated by mixedmath enough to show it?
@GustavoBandeira Hi.
@GustavoBandeira What they want you to do is to show that the last claim (about powers) implies the claim about the polynomials.
Yes. But how do I do that?
"Since they agree on kth powers, and polynomials are sums of kth powers..."
18:23
@PedroTamaroff Oh. I'll think about it
18:41
@PedroTamaroff I get$$\frac{\pi}{\sqrt{1-2r-3r^2}} \left(\left(\frac{1-r-\sqrt{1-2r-3r^2}}{2r^2}\right)^n + \left(\frac{1-r+\sqrt{1-2r-3r^2}}{2r^2}\right)^n\right)$$
It looks like I was too late...
@Chris'ssis Did you verify Pedro's answer?
@robjohn Yes, it seems correct. What did you use to get that answer?
@Chris'ssis Contour integration
I will check numerically
@robjohn Ah, I thought you got it by real methods.
@robjohn Pedro's answer seems correct.
@robjohn You and your fancy complex analysis!
18:54
I am really unsure about what subjects to take next semester =/
@N3buchadnezzar What have you taken? What can you take?
@N3buchadnezzar pick them randomly
@PedroTamaroff I have taken most introductory courses, basic number theory, basic functional analysis, some abstract algebra and some real analysis. Also a fair bit of linear algebra.
My choices are many.. Galois theory, Manifolds, Complex Analysis, General Topology, Analytic Numbertheory, Classical Mechanics etc.
@N3buchadnezzar Well, interested in any of those?
00:00 - 19:0019:00 - 00:00

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